>Butch Malahide wrote:>>Butch Malahide wrote:>>>Butch Malahide wrote:>>>>quasi wrote:>>>>>Butch Malahide wrote:>>>>>>quasi wrote:>>>>>>>quasi wrote:>>>>>>>>quasi wrote:>>>>>>>>>quasi wrote:>>>>>>>>>>Butch Malahide wrote:>>>>>>>>>>>David C. Ullrich wrote:>>>>>>>>>>>>Butch Malahide wrote>>>>>>>>>>>>>William Elliot wrote:>>>>>>>>>>>>>>David Hartley wrote:>>>>>>>>>>>>>>>William Elliot wrote:>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>Perhaps you could illustrate with the five>>>>>>>>>>>>>>>>different one to four point point>>>>>>>>>>>>>>>>compactifications of two open end line>>>>>>>>>>>>>>>>segements.>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>(There are seven.)>>>>>>>>>>>>>>>>>>>>>>>>>>>>Ok, seven non-homeomophic finite Hausdorff>>>>>>>>>>>>>>compactications.>>>>>>>>>>>>>>>>>>>>>>>>>>How many will there be if you start with n segments>>>>>>>>>>>>>instead of 2?>>>>>>>>>>>>>>>>>>>>>>>>Surely there's no simple formula for that?>>>>>>>>>>>>>>>>>>>>>>>> ...>>>>>>>>>>>>>>>>>>>>>> ...>>>>>>>>>>>>>>>>>>>>>>I wasn't necessarily expecting a *complete* answer, >>>>>>>>>>>such as an explicit generating function. Maybe someone>>>>>>>>>>>could give a partial answer, such as an asymptotic>>>>>>>>>>>formula, or nontrivial upper and lower bounds, or a>>>>>>>>>>>reference to a table of small values, or the ID number>>>>>>>>>>>in the Encyclopedia of Integer Sequences, or just the >>>>>>>>>>>value for n = 3. (I got 21 from a hurried hand count.)>>>>>>>>>>>>>>>>>>>>For n = 3, my hand count yields 19 distinct>>>>>>>>>>compactifications, up to homeomorphism.>>>>>>>>>>>>>>>>>>>>Perhaps I missed some cases.>>>>>>>>>>>>>>>>>>>I found 1 more case.>>>>>>>>>>>>>>>>>>My count is now 20.>>>>>>>>>>>>>>>>I found still 1 more case.>>>>>>>>>>>>>>>>So 21 it is!>>>>>>>>>>>>>>>>But after that, there are no more -- I'm certain.>>>>>>>>>>>>>>Oops -- the last one I found was bogus.>>>>>>>>>>>>>>So my count is back to 20.>>>>>>>>>>>>Hmm. I counted them again, and I still get 21.>>>>>>>>>>>>4 3-component spaces: OOO, OO|, O||, |||.>>>>>>>>>>>>7 2-component spaces: OO, O|, O6, O8, ||, |6, |8.>>>>>>>>>>>>10 connected spaces: O, |, 6, 8, Y, theta, dumbbell, and>>>>>>the spaces obtained by taking a Y and gluing one, two, or>>>>>>all three of the endpoints to the central node.>>>>>>>>>>Thanks.>>>>>>>>>>It appears I missed the plain "Y", but other than that,>>>>>everything matches.>>>>>>>>>>So yes, 21 distinct types.>>>>>>>> For n = 4 I get 56 types. If I counted right (very iffy), >>>>>> Found two more. Never mind!>>>>And now I get 61. The hell with it.>>Ullrich predicted it (hopeless squared).

Hey, you guys are making progress!Now all we need is an f such thatf(2) = 7, f(3) = 21 or whatever it was,and f(4) = 61. I doubt there's more thanone such f...

>>For small n, say n < 10, it might be feasible to get the counts>via a computer program, but my sense is that the development of >such a program would be fairly challenging. If I get a chance,>I may give it a try.

My prediiction is that anything remotely resembling bruteforce will take forever to run, even in a compiled language.Just enumerating all the possible equivalence relationson 2n points is going to take a long time; now abrute force check whether _one_ pair of equivalencerelations defines homeomorphic spaces will also takea long time, and there will be N(N-1)/2 such checks,where N is huge even for small n.

But don't let me stop you. I'd start in Python, whichwould be much too slow - the point being that Pythonis so easy to _write_ that it would be the quickestway to get the algorithm straight. I'd test thePython code for n = 2 and 3 maybe; for n = 3it would take forever. Once I had a correctalgorithm I'd rewrite it in a compiled language.Then ask for the answer for n = 4 and maybetake a trip somewhere while waiting for theanswer...